X>0 Y>0证1/x+1/y>=4/(x+y)

若|x+2y-1|+y²+4y+4=0,求（2x-y）²-2（2x-y)(x+2y)+(x+2y)& x>1,y>0,且满足xy=x^y,x/y=x^3y,求x+y 设x>1,y>0,若x^y+x^-y=2根号2,则x^y-x^-y等于 x>0 y>0 x+y=1,4/X+9/y最小值 实数x,y满足{x-y+1>=0;y+1>=0;x+y+1 以下三个行列式的计算过程是什么.谢谢 1.x y x+y y x+y x x+y x y 2.1 x y z x 1 0 微分方程(1)(y'')^2+5y'-y+x=0;(2)y''+5y'+4y^2-8x=0;(3)(3x+2y)dx+( 若2x-3y+4=0则x（x*x-1)+x(5-x*x)-6y+7 若实数x,y满足x-y+1>=0,x+y>=0,x 若实数x,y满足x-y+1》=0,x+y>=0,x 已知x,y满足约束条件：x-y+1>=0,x+y-2>=0,x 若|x+y-1|+（x-y-2)²=0,求代数式（x+2y)(x-2y）-（2x-y)(-y-2x)的值.